This photo and problem is taken from Zee's textbook pg. 169, Einstein's Gravity in a nutshell. Here in the moving reference frame we're moving at a velocity $u$, and the total time it takes to go from the top mirror back to the bottom mirror is given by ($\Delta t$ in the rest frame and $\Delta t'$ in the moving frame). The total displacement in the rest frame is $\Delta x=0$ since the initial and final position are the same, but in the moving frame this isn't the case. That implies (1) $c\Delta t' = 2\sqrt{(\frac{1}{2}u\Delta t')^2 + L^2}$ is the distance light travels. We also know $\Delta x' = u\Delta t'$.
Problem 1 in this section says directly from equation (1) we can get the lorentz transformations, which are:
$ct' =\frac{ct+\frac{u}{c}x}{\sqrt{1-\frac{u^2}{c^2}}}$
$x' = \frac{x+\frac{u}{c}ct}{\sqrt{1-\frac{u^2}{c^2}}}$
Also: $y'=y$ & $z'=z$
We also know $2L/c = \Delta t$.
Every time I apply this formula I get $\Delta t' = \gamma \Delta t$, where $\gamma = (1-\frac{v^2}{c^2})^{-1/2}$, but this I cannot match with the lorentz transf. equations. I also believe possibly $\Delta t' = t' -t$ & $\Delta x' = x' - x$ but I do not know for certain and this hasn't helped me at all.